IB Math AA SL Topic 3.8: Trig equations | Notes & Flashcards | Aimnova

IB Math AA SL — Trig equations

Topic 3.8 of IB Mathematics: Analysis and Approaches covers Trig equations, which is part of Unit 3: Geometry & Trigonometry. Students explore key concepts including Solving trig equations. A strong understanding of trig equations is essential for IB Math AA SL exams and builds the foundation for connected topics across the syllabus.

Key concepts in Trig equations

Key Idea: Trig equations ask you to find every angle in a given domain that satisfies the equation — and because sin, cos and tan repeat, there is almost always more than one answer. Worth full marks on both papers.

🎯 The method: base angle, then all its partners

Over one full turn, sin x = k and cos x = k (with |k| < 1) each give two solutions; tan x = k gives one per 180°. If k is negative, the angles sit in different quadrants — sketch the unit circle (or use CAST) to place them, e.g. cos x = −0.5 lands in Q2 and Q3.

🔁 Two cases that hide extra answers

✏️ IB-style worked examples

IB-style question — find all solutions in the domain (Paper 1)

Solve cos x = 0.5 for 0° ≤ x ≤ 360°, without a calculator.

Step by step:

Take the inverse for the base angle.
$x = \cos^{-1}(0.5) = 60^\circ$
Second solution for cosine: 360° − x.
$360^\circ - 60^\circ = 300^\circ$
Both lie in 0° ≤ x ≤ 360°, so keep both.
$x = 60^\circ,\; 300^\circ$

Final answer:

x = 60° or x = 300°.

IB-style question — a multiple-angle equation (Paper 1)

Solve sin(2x) = 0.5 for 0° ≤ x ≤ 360°.

Step by step:

Let u = 2x, so the interval stretches to 0° ≤ u ≤ 720°.
$\sin u = 0.5$
All u in the doubled interval (base 30°, then 180°−30°, then +360°).
$u = 30^\circ,\;150^\circ,\;390^\circ,\;510^\circ$
Divide each by 2 to get x.
$x = 15^\circ,\;75^\circ,\;195^\circ,\;255^\circ$

Final answer:

x = 15°, 75°, 195°, 255° (four solutions).

IB-style question — a quadratic in sin x (Paper 1)

Solve 2sin²x − sin x − 1 = 0 for 0° ≤ x ≤ 360°.

Step by step:

Let s = sin x and factor the quadratic.
$(2s + 1)(s - 1) = 0$
So sin x = 1 or sin x = −½ (both are in [−1, 1]).
$\sin x = 1 \;\text{or}\; \sin x = -\tfrac{1}{2}$
Solve each over the interval.
$x = 90^\circ;\quad x = 210^\circ,\;330^\circ$

Final answer:

x = 90°, 210°, 330°.

IB-style question — solve inside a model with the GDC (Paper 2)

A tide height is h = 4 sin(30t)° + 6 (t in hours). Find the first time t > 0 when the height is 8 m.

Step by step:

Set up the equation.
$4\sin(30t)^\circ + 6 = 8 \Rightarrow \sin(30t)^\circ = 0.5$
Graph y = h and y = 8 on the GDC and use intersect over the interval.
$30t = 30^\circ$
Solve for the first time.
$t = 1\ \text{hour}$

Final answer:

First at t = 1 hour (the GDC intersect confirms it).

Important: The inverse on your calculator gives only one angle. The domain almost always wants more — find the partner (sin → 180°−x, cos → 360°−x, tan → +180°) and check it's inside the interval. For a multiple angle, find every solution in the stretched interval before you divide.

Tap each card to reveal the answer.

Exam Tips

Underline the domain first — it decides how many answers you keep.
Inverse gives one angle; always find the partner (sin → 180°−x, cos → 360°−x, tan → +180°).
Multiple angle: solve over the stretched interval, then divide — don't halve too early.
Quadratic in sin/cos: substitute s = the ratio, factor, and reject any s outside [−1, 1].
Paper 2: graph both sides and use intersect — perfect for 'when is the height …?' models.

IB Math AA SL — Trig equations

Key concepts in Trig equations

Key Idea: Trig equations ask you to find every angle in a given domain that satisfies the equation — and because sin, cos and tan repeat, there is almost always more than one answer. Worth full marks on both papers.

🎯 The method: base angle, then all its partners

Over one full turn, sin x = k and cos x = k (with |k| < 1) each give two solutions; tan x = k gives one per 180°. If k is negative, the angles sit in different quadrants — sketch the unit circle (or use CAST) to place them, e.g. cos x = −0.5 lands in Q2 and Q3.

🔁 Two cases that hide extra answers

✏️ IB-style worked examples

IB-style question — find all solutions in the domain (Paper 1)

Solve cos x = 0.5 for 0° ≤ x ≤ 360°, without a calculator.

Step by step:

Take the inverse for the base angle.
$x = \cos^{-1}(0.5) = 60^\circ$
Second solution for cosine: 360° − x.
$360^\circ - 60^\circ = 300^\circ$
Both lie in 0° ≤ x ≤ 360°, so keep both.
$x = 60^\circ,\; 300^\circ$

Final answer:

x = 60° or x = 300°.

IB-style question — a multiple-angle equation (Paper 1)

Solve sin(2x) = 0.5 for 0° ≤ x ≤ 360°.

Step by step:

Let u = 2x, so the interval stretches to 0° ≤ u ≤ 720°.
$\sin u = 0.5$
All u in the doubled interval (base 30°, then 180°−30°, then +360°).
$u = 30^\circ,\;150^\circ,\;390^\circ,\;510^\circ$
Divide each by 2 to get x.
$x = 15^\circ,\;75^\circ,\;195^\circ,\;255^\circ$

Final answer:

x = 15°, 75°, 195°, 255° (four solutions).

IB-style question — a quadratic in sin x (Paper 1)

Solve 2sin²x − sin x − 1 = 0 for 0° ≤ x ≤ 360°.

Step by step:

Let s = sin x and factor the quadratic.
$(2s + 1)(s - 1) = 0$
So sin x = 1 or sin x = −½ (both are in [−1, 1]).
$\sin x = 1 \;\text{or}\; \sin x = -\tfrac{1}{2}$
Solve each over the interval.
$x = 90^\circ;\quad x = 210^\circ,\;330^\circ$

Final answer:

x = 90°, 210°, 330°.

IB-style question — solve inside a model with the GDC (Paper 2)

A tide height is h = 4 sin(30t)° + 6 (t in hours). Find the first time t > 0 when the height is 8 m.

Step by step:

Set up the equation.
$4\sin(30t)^\circ + 6 = 8 \Rightarrow \sin(30t)^\circ = 0.5$
Graph y = h and y = 8 on the GDC and use intersect over the interval.
$30t = 30^\circ$
Solve for the first time.
$t = 1\ \text{hour}$

Final answer:

First at t = 1 hour (the GDC intersect confirms it).

Important: The inverse on your calculator gives only one angle. The domain almost always wants more — find the partner (sin → 180°−x, cos → 360°−x, tan → +180°) and check it's inside the interval. For a multiple angle, find every solution in the stretched interval before you divide.

Tap each card to reveal the answer.

Exam Tips

Underline the domain first — it decides how many answers you keep.
Inverse gives one angle; always find the partner (sin → 180°−x, cos → 360°−x, tan → +180°).
Multiple angle: solve over the stretched interval, then divide — don't halve too early.
Quadratic in sin/cos: substitute s = the ratio, factor, and reject any s outside [−1, 1].
Paper 2: graph both sides and use intersect — perfect for 'when is the height …?' models.

IB Math AA SL — Trig equations

Key concepts in Trig equations

🎯 The method: base angle, then all its partners

🔁 Two cases that hide extra answers

✏️ IB-style worked examples

IB-style question — find all solutions in the domain (Paper 1)

IB-style question — a multiple-angle equation (Paper 1)

IB-style question — a quadratic in sin x (Paper 1)

IB-style question — solve inside a model with the GDC (Paper 2)

Exam Tips

What you'll learn in Topic 3.8

Study resources — 3.8 Trig equations

Solving trig equations

Ready to study Trig equations?

Ready to practice?

IB Math AA SL — Trig equations

Key concepts in Trig equations

🎯 The method: base angle, then all its partners

🔁 Two cases that hide extra answers

✏️ IB-style worked examples

IB-style question — find all solutions in the domain (Paper 1)

IB-style question — a multiple-angle equation (Paper 1)

IB-style question — a quadratic in sin x (Paper 1)

IB-style question — solve inside a model with the GDC (Paper 2)

Exam Tips

What you'll learn in Topic 3.8

Study resources — 3.8 Trig equations

Solving trig equations

Ready to study Trig equations?

Ready to practice?